The area of the region bounded by the lines \(\displaystyle\frac{x}{7\sqrt{3}\,a}+\frac{y}{b}=4\), x = 0 and y = 0 is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(56\sqrt{3}\,ab\)
The region is the right-angled triangle formed by the line and the two axes. Dividing the equation by 4 puts it in intercept form:
\( \displaystyle \frac{x}{28\sqrt{3}\,a} + \frac{y}{4b} = 1 \)
So the line cuts the x-axis at \(28\sqrt{3}\,a\) and the y-axis at 4b (taking a, b > 0). The area of the triangle is
\( \displaystyle A = \frac{1}{2}\times 28\sqrt{3}\,a \times 4b \)
\( = 56\sqrt{3}\,ab \)